= Solution
Use $\hbar=c=1$ and the mostly-plus <Minkowski metric>. Then the source's <Klein-Gordon equation> is $(\partial_t^2-\nabla^2+m^2)\phi=0$. For a <plane wave> $\phi=C e^{-i\omega t+i\mathbf k\cdot\mathbf x}$, substitution gives
$$
\boxed{\omega^2=|\mathbf k|^2+m^2.}
$$
Thus a real propagating <wave vector> requires $|\omega|\ge m$, or $|E|\ge mc^2$ on restoring units. The positive-<energy> branch has a rest-energy threshold $mc^2$. At $|\omega|=m$, $\mathbf k=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real <scalar field> combines complex-conjugate modes. Its <group velocity> is $\mathbf k/\omega$ and is subluminal, though the positive-branch phase <speed> exceeds one when $m>0$.
For a static spherically symmetric <scalar field>, the equation outside the source is
$$
\frac1{r^2}\frac d{dr}\left(r^2\frac{d\phi}{dr}\right)-m^2\phi=0.
$$
Setting $\psi=r\phi$ gives $\psi''-m^2\psi=0$, so
$$
\boxed{\phi(r)=\frac{A e^{-mr}+B e^{mr}}r\quad(r>0).}
$$
Decay at infinity selects $B=0$, the <Yukawa potential>. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\phi=A/r+B$, and decay sets the constant $B$ to zero.
Back to article page
